我使用链表编写了以下队列实现,该链表不维护对尾节点的引用。当我尝试打印队列时,它仅输出头部,即仅输出一个节点。错误是什么?提前致谢!
package DataStructures;
import java.util.Scanner;
class Node {
int x;
Node nextNode;
public Node(int x) {
this.x = x;
nextNode = null;
}
}
class Queue {
Node head = null;
int n = 0;
public void enqueue(int x) {
if (n==0){
head = new Node(x);
n++;
return;
}
Node tempHead = head;
while (tempHead != null){
tempHead = tempHead.nextNode;
}
tempHead = new Node(x);
tempHead.nextNode = null;
n++;
}
public int dequeue() {
if (head == null) {
throw new Error("Queue under flow Error!");
} else {
int x = head.x;
head = head.nextNode;
return x;
}
}
public void printTheQueue() {
Node tempNode = head;
System.out.println("hi");
while (tempNode != null){
System.out.print(tempNode.x + " ");
tempNode = tempNode.nextNode;
}
}
}
public class QueueTest {
private static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
Queue queue = new Queue();
while (true){
int x = in.nextInt();
if (x == -1){
break;
} else{
queue.enqueue(x);
}
}
queue.printTheQueue();
}
}
最佳答案
您永远不会将节点分配给 nextNode
,因此您的列表要么为空,要么由一个节点组成。
这是一个解决方案:
public void enqueue(int x) {
n++;
if (head == null) {
head = new Node(x);
else {
Node last = head;
while (last.nextNode != null)
last = last.nextNode;
last.nextNode = new Node(x);
}
}
从技术上讲,您不需要n
,但您可以将其用作列表大小的缓存。并且您应该在 deque()
中减少它。
关于java - 下面使用链表实现队列有什么错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32023989/